anybody know how to integrate tanx/cosx?

Answer 1

tanx/cosx = sinx/cos^2x,

let u = cosx , you have du = -sinx dx, => sinxdx = -u,

plug into the original eq, the integral become -1/u^2 du, the result
is 1/u ie, 1/cosx + C

Answer 2

Let me try to remember some of my old calculus.

since tanx = sinx/cosx

tanx/cosx = sinx/(cosx)^2

let u = cosx then du = -sinx

you now have an integral in the form:

- integral of du/u^2 = - integral of u^(-2) du

which equals - [u^(-1)]/(-1) = u^(-1) = 1/u + constant

substituting back we get 1/cosx + C

Answer 3

in a circle tan = y/x and cos = x/r (r = radius) thus y/x * r/x = yr/x^2 idk if thats what you want